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3 years ago

At what location does gravity play a role in moving tectonic plates

Physics

2 answers:

Ksju [112]3 years ago

8 0

At the edge of the plates.

Lena [83]3 years ago

7 0

Answer:

In subduction of one plate over another on the basis of weight

Explanation:

When two plates moving and colliding, the plate with higher mass will subduct below the plate with lower weigh and lighter would remain above. generally continental types of plates are lighter in weight and oceanic plates are heavier.

Ex:- The collision of Eurasian plate and Indian plate. Indian plate is below and Eurasian plate above and formation of Himalaya

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The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

3 years ago

Commercials for a toy bouncy ball advertise that it will bounce to a height that is greater than the

storchak [24]

Answer:

because energy will be lost due to friction, sound, and heat (arguably similar to friction) and ENERGY MUST STAY THE SAME so it is IMPOSSIBLE for the ball to bounce higher than when dropped!

7 0

3 years ago

How is social media changing the world as we know it?

Vika [28.1K]

Answer:

Social media has helped many businesses grow and promote itself, and has helped people find a better way to connect and communicate with one another. On the other hand, it's also provided many people with problems involving mental health, emotional insecurities, and waste of time

Explanation:

what they said is correct

4 0

2 years ago

The wavelength of light that has a frequency of 1.20 × 1013 s-1 is ________ m.

klio [65]

The relationship between frequency and wavelength for an electromagnetic wave is
$c=f \lambda$
where
f is the frequency
$\lambda$ is the wavelength
$c=3 \cdot 10^8 m/s$ is the speed of light.

For the light in our problem, the frequency is $f=1.20 \cdot 10^{13} s^{-1}$ , so its wavelength is (re-arranging the previous formula)
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{1.20 \cdot 10^{13} s^{-1}}= 2.5 \cdot 10^{-5}m$

8 0

3 years ago

A 25kg chair initially at rest on a horizontal floor requires a 165 N horizontal force to set it in motion. Once the chair is in

SCORPION-xisa [38]

Answer:

$\mu_k=0.51$

Explanation:

Given that

Mass , m = 25 kg

We know that when body is in rest condition then static friction force act on the body and when body is in motion the kinetic friction force act on the body .That is why these two forces are given as follows

Static friction force ,fs= 165 N

Kinetic friction force ,fk = 127 N

If the body is moving with constant velocity ,it means that acceleration of that body is zero and all the forces are balanced.

Lets take coefficient of kinetic friction = μk

The kinetic friction is given as follows

fk = μk m g

Now by putting the values

127 = μk x 25 x 9.81

$\mu_k=\dfrac{127}{25\times 9.81}$

$\mu_k=0.51$

Therefore the value of coefficient of kinetic friction will be 0.51

4 0

3 years ago